In the given diagram, ABC is a triangle, where B(4, −4) and C(−4, −2). D is a point on AC.
(a) Write down the coordinates of A and D.
(b) Find the coordinates of the centroid of △ABC.
(c) If D divides AC in the ratio k : 1, find the value of k.
(d) Find the equation of the line BD.
(a) From the graph:
A = (0, 6) and D = (−3, 0).
(b) The centroid of a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is:
\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)For A(0, 6), B(4, −4) and C(−4, −2):
Centroid = \left(\frac{0+4-4}{3},\frac{6-4-2}{3}\right)
⇒ Centroid = (0, 0)
(c) D divides AC in the ratio k : 1.
Using the section formula:
D=\left(\frac{k(-4)+1(0)}{k+1},\frac{k(-2)+1(6)}{k+1}\right)Since D = (−3, 0):
-3=\frac{-4k}{k+1}⇒ −3(k + 1) = −4k
⇒ −3k − 3 = −4k
⇒ k = 3
Checking the y-coordinate:
0=\frac{-2k+6}{k+1}⇒ −2k + 6 = 0
⇒ k = 3
(d) The line BD passes through B(4, −4) and D(−3, 0).
Slope = \frac{0-(-4)}{-3-4}=\frac{4}{-7}=-\frac{4}{7}
Using the point-slope form with B(4, −4):
y − (−4) = -\frac{4}{7}(x − 4)
⇒ 7(y + 4) = −4(x − 4)
⇒ 7y + 28 = −4x + 16
⇒ 4x + 7y + 12 = 0
Therefore, the equation of BD is 4x + 7y + 12 = 0.
More ICSE Hard level questions
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